Solid ammonium carbamate $(N H_{2} C O O N H_{4})$ is in equilibrium with $N H_{3} (g)$ and $C O_{2} (g)$ in a closed vessel at a certain temperature T(K). The total pressure at equilibrium is 27 atm ammonia gas is now introduced into the system at equilibrium till the partial pressure of $N H_{3}$ at new equilibrium is equal to the original total pressure. What is the total pressure (in atm) at new equilibrium?

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Detailed Solution

$N H_{2} C O O N {H_{4}}_{(s)} ⇄ 2 N H_{3 (g)} + C O_{2 (g)}$

Pressure at 2p p
equilibrium
$2 p + p = 27 \Rightarrow p = 9 atm \Rightarrow K_{p} = {(P_{N H_{3}})}^{2} (P_{C O_{2}}) = {(18)}^{2} \times 9 {atm}^{3}$
Partial pressure of NH₃ at new equilibrium = 27 atm
$K_{p} = {(P_{N H_{2}}^{1})}^{2} \times P_{C O_{2}}^{1} {(18)}^{2} \times 9 = {(27)}^{2} . p_{C O_{2}}^{1} P_{C O_{2}}^{1} = {(\frac{18}{27})}^{2} \times 9 = 4$